2021-06-20 · 1분
BOJ 11047 — Coin 0
A short write-up for BOJ 11047 (Coin 0), a greedy coin problem.
algorithmboj
2021-06-16 · 2분
This post is over 2 years old. The content may be outdated.
https://www.acmicpc.net/problem/7569 <- the problem-solving site
#include <cstdio>
#include <iostream>
#include <queue>
#include <tuple>
#define endly "\n"
using namespace std;
int d[100][100][100];
bool visited[100][100][100];
int dx[] = {1, -1, 0, 0, 0, 0};
int dy[] = {0, 0, 1, -1, 0, 0};
int dz[] = {0, 0, 0, 0, 1, -1};
queue<tuple<int, int, int>> q;
int main() {
int m, n, h;
cin >> m >> n >> h;
bool all_one;
for (int z = 0; z < h; z++) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> d[i][j][z];
// check whether 0 was entered
if (d[i][j][z] == 0) all_one = true;
if (d[i][j][z] == 1) {
q.push(make_tuple(i, j, z));
visited[i][j][z] = true;
}
}
}
}
// if there is no input of 0
if (!all_one) {
printf("0");
return 0;
}
while (!q.empty()) {
int x, y, z;
tie(x, y, z) = q.front();
q.pop();
for (int k = 0; k < 6; k++) {
int nx = x + dx[k];
int ny = y + dy[k];
int nz = z + dz[k];
if (nx >= 0 && nx < n && ny >= 0 && ny < m && nz >= 0 && nz < h) {
if (d[nx][ny][nz] == 0 && !visited[nx][ny][nz]) {
q.push(make_tuple(nx, ny, nz));
visited[nx][ny][nz] = true;
d[nx][ny][nz] = d[x][y][z] + 1;
}
}
}
}
int ans = 0;
for (int z = 0; z < h; z++) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (d[i][j][z] == 0) { // if there is an unripe tomato
printf("-1");
return 0;
}
if (ans < d[i][j][z]) ans = d[i][j][z];
}
}
}
printf("%d", ans-1);
return 0;
}
Original (Korean): tistory — published 2021-06-16, migrated to this blog. This translation was generated with the help of AI.
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